chapter 2 wave nature of matter (pp 43-72)

CH 02 WAVE NATURE OF MATTER 43

CHAPTER 2 WAVE NATURE OF MATTER

2-1 DE BROGLIE WAVELENGTH Problem 2-1 A bullet of mass 41 g travels at 960 m s-1. (a) What wavelength can we associate with it? (b) Why does the wave nature of the bullet not reveal itself through the diffraction effects? Solution (a) The de Broglie wavelength is given by m

mv

h 353

34

10683.1)960)(1041()10626.6(

=

==

(b) This value of wavelength is extremely small as compared with any reasonable microscopic distance, therefore the wave nature of the bullet is not revealed through diffraction effects.

Problem 2-2 What will be the de Broglie wavelength associated with a mass of 0.01 kg moving with a velocity of 10 m/s? K.U. B.Sc. 2002 Solution The de Broglie wavelength associated with a moving object is given by

mmv

h 3334 10626.6)10)(01.0(10626.6

=

==

Problem 2-3 Calculate the de Broglie wavelength of virus particle of mass 15101 kg moving at a speed of 2 mm/s. P.U. B.Sc. 2005


Solution The de Broglie wavelength is given by

mmv

h 16315

34

10313.3)102)(101()10626.6(

=

==

Problem 2-4 A tennis ball having mass 20 gm is moving at a speed of 50 m/s. Compare its de Broglie wavelength to that of an electron moving with a velocity of 6105 m/s. Solution The de Broglie wavelength associated with a moving object is given by

vm

h=

For tennis ball

mvm

h 343

34

10626.6)50)(1020()10626.6(

=

==

This wavelength is too small to be observed experimentally. For de Broglie wavelength of electron we have

mvm

he

10631

34

0

10455.1)105)(10109.9()10626.6(

=

==

Now 241034

10554.410455.110626.6

=

=

e

Problem 2-5 Find the de Broglie wavelength of 1.0 mg grain of sand blown by the wind at a speed of 20 m/s. Solution The de Broglie wavelength is given by

mvm

h 296

34

10313.3)20)(100.1(10626.6

=

==


Problem 2-6 Calculate the de Broglie wavelength of an electron moving with a speed of 9 x 106 m s -1. B.U. B.Sc. 2002A Solution The de Broglie wavelength is given by

mvm

h 11631

34

0

10082.8)109)(10109.9()10626.6(

=

==

= 808.0 Problem 2-7 The speed of electron in first orbit of Hydrogen atom is

6102 m/s. Calculate the de Broglie wavelength. Solution The de Broglie wavelength is given by

)102)(10109.9()10626.6(

631

34

0

==

vm

h

64.31064.3 10 == m This wavelength is comparable to the size of atom and should be observable. Problem 2-8 If the de Broglie wavelength of an electron is 10101.1 m, what is the speed of the electron? P.U. B.Sc. 2006 Solution The de Broglie wavelength is given by

vm

h0

=

smm

hv /10613.6)101.1)(10109.9(

)10626.6( 61031

34

0

=

==

Problem 2-9 What will be the speed of an electron with an associated wavelength of 2.00 ? Solution The de Broglie wavelength is defined as


vm

hph

0

==

0mh

v =

smv /1064.3)1000.2)(10109.9()10626.6( 6

1031

34

=

=

Problem 2-10 Calculate the de Broglie wavelength of a 1.00 keV (a) electron, (b) neutron and (c) a photon. Solution Now Kmv =2

21

Kmv 22 = mKvm 222 = or mKp 22 = or mKp 2=

mKh

ph

2==

(a) The de Broglie wavelength for an electron is given by

)10602.1101)(10109.9(210626.6

2 1933134

==

Kmh

e

= m111088.3 or 38.8 pm

(b) For a neutron m is replaced by mn i.e.

)10602.1101)(10675.1(210626.6

2 1932734

==

Kmh

n

= m131004.9 or 904 fm

(a) The energy of a photon is given by relation

hcE =

or )10602.1101()10998.2)(10626.6(

193

834

==

Ehc

m91024.1 = or 1.24 nm


Problem 2-11 Calculate the wavelength associated with an alpha particle emitted by the nucleus of an atom of Radon-226. The kinetic energy of alpha particle emitted from Radon-222 is 5.486 MeV. Solution The de Broglie wavelength corresponding to this energy is

Kmh

2

=

)10602.110486.5)(10445.6(210626.6

19627

34

=

Q JeV 1910602.11 = m

1510225.6 =

Problem 2-12 Find the de Broglie wavelength of a 40 keV electrons used in a certain electron microscope. Solution The de Broglie wavelength corresponding to kinetic energy K is

)10602.11050)(10109.9(210626.6

2 1933134

0

==

Kmh

m1210133.6 =

Problem 2-13 Find the wavelength of an electron which has been accelerated by a potential difference of 54 volts. B.U. B.Sc. 1989A, 1992A Solution Now eVvmEK 0

202

1.. ==

eVvm 02

0 2= or eVmvm 0022

0 2=


eVmvmp 000 2== The de Broglie wavelength is given by

)10602.1)(54)(10109.9(2)10626.6(

2 191934

00

===

eVmh

ph

m1010669.1 = or 1.669

Problem 2-14 Suppose we are to build an electron microscope and we want to operate at a wavelength of 0.10 nm. What accelerating should be used? Solution Now 200 2

1vmeV =

em

hem

pem

vm

e

vmV0

2

0

2

0

20

20

0 2)/(

222

====

20

2

0 2 emhV =

voltsV 150)1010.0)(10602.1)(10109.9(2)10626.6(

291931

234

0 =

=

Problem 2-15 Green light has a wavelength of about 550 nm. Through what potential difference must an electron be accelerated to have this wavelength? Solution Now 200 2

1vmeV =

em

hem

pem

vm

e

vmV0

2

0

2

0

20

20

0 2)/(

222

====

20

2

0 2 emhV =


291931

234

0 )10550)(10602.1)(10109.9(2)10626.6(

=V

VvoltsV 97.41097.4 60 == Problem 2-16 Calculate the de Broglie wavelength of 5 MeV alpha particles. P.U. B.Sc. 1990 Solution As given the energy of alpha particle, 5 MeV, is much smaller than its rest mass energy i.e. MeVcm 37272 = , therefore it is a non-relativistic case. The de Broglie wavelength is given by

Kmh

ph

2

==

where Jsh 3410626.6 = , kgm 2710645.6 = and JJeVMeVK 131966 1001.8)10602,1)(105(1055 ====

Hence

fmm 836.610836.6)1001.8)(10645.6(2

10626.6 151327

34

==

=

Problem 2-17 A proton and electron have the same de Broglie wavelength. How does their speeds compare, assuming both are much less than that of light? Solution Now ep =

eepp vm

hvm

h=

331

27

10836.110109.910673.1

=

==

e

p

p

e

m

m

v

v

The velocity of electron is 1836 times the velocity of proton.


Problem 2-18 Calculate the ratio of the kinetic energy of an electron to that of a proton if their wavelengths are equal. Assume that the speeds are nonrelativistic. Solution The kinetic energy of a particle is given by

2

222222

22)/(

2221

m

hm

hm

pm

vmvmK =====

Now

2

2

2 ee

m

hK =

2

2

2 pp

m

hK = Q the wavelengths are equal.

Therefore e

p

p

e

p

e

m

m

mhmh

KK

== )2/()2/(

22

22

183710109.910673.1

31

27

=

=

p

e

KK

Problem 2-19 Find the kinetic energy of an electron whose de Broglie wavelength is the same as that of 100 keV X-ray. B.U. B.Sc. 2009S Solution The energy of the given X-ray photon is

JeVkeVE )10602.1)(101(10100100 1953 === JE 1410602.1 =

But chE =

mE

ch 1114

834

10239.110602.1

)10998.2)(10626.6(

=

==

This is the de Broglie wavelength of given electron. The kinetic energy of the electron is given by


20

2

0

2

0

2

0

2202

0 22)/(

2221

m

hm

hm

pm

vmvmK =====

JK 1521131234

10570.1)10239.1)(10109.9(2)10626.6(

=

=

keVeVeVK 80.91080.910602.110570.1 3

19

15

==

=

Problem 2-20 Calculate the kinetic energy of a proton having a de Broglie wavelength of 0.4 . Solution The kinetic energy of the proton is given by

pp

pp

m

pm

vmvmK

2221 2222

===

2

22

22)/(

pp m

hm

hK == hp =Q

JK 2021027234

10201.8)104.0)(10673.1(2)10626.6(

=

=

eVeVK 512.010602.110201.8

19

20

=

=

Problem 2-21 Calculate the de Broglie wavelength of an electron whose kinetic energy is 120 eV. P.U. B.Sc. 2000, 2003, F.P.S.C. 2007, K.U. B.Sc. 2007 Solution The de Broglie wavelength is given by

Kmh

ph

02==

)10602.1120)(10109.9(210626.6

1931

34

=


Q JeV 1910602.11 = m

101012.1 = or 1.12

Problem 2-22 Calculate the de Broglie wavelength of electrons with kinetic energy 500 eV. Solution The de Broglie wavelength of electrons with kinetic energy K is given by

Kmh

ph

02==

)10602.1500)(10109.9(210626.6

1931

34

=

nmm 055.010485.5 11 ==

Problem 2-23 Calculate the de Broglie wavelength of an electron whose kinetic energy is 1 keV. P.U. B.Sc. 2007 Solution Now Kvm =202

1

Kvm 220 = Kmvm 0

220 2=

Kmp 02 2= or Kmp 02=

Kmh

ph

02==

)10602.1101)(10109.9(210626.6

19331

34

=

pmm 8.381088.3 11 ==


Problem 2-24 Find the kinetic energy of an electron with the same wavelength as blue light ( = 450 nm). Solution The kinetic energy of a particle is given by

2

222222

22)/(

2221

m

hm

hm

pm

vmvmK =====

Now

2

2

2 ee

m

hK =

JKe24

2931

234

10190.1)10450)(10109.9(2)10626.6(

=

=

eVeVKe6

19

24

1043.710602.110190.1

=

=

Problem 2-25 Find the kinetic energy of a neutron with the same wavelength as blue light ( = 450 nm). Solution The kinetic energy of a particle is given by

2

222222

22)/(

2221

m

hm

hm

pm

vmvmK =====

Now 22

2 nn mhK =

JKe28

2927

234

10472.6)10450)(10673.1(2)10626.6(

=

=

eVeVKe9

19

28

1004.410602.110472.6

=

=

Problem 2-26 Find the de Broglie wavelength of electron, proton and alpha particle all having same kinetic energy of 100 eV.


Solution The de Broglie wavelength in terms of kinetic energy K is given by

mKh

ph

2==

mm

25

19

34 10171.1)10602.1100(2

10626.6(

=

=

For electron kgmm e3110109.9 == and

31

2525

10109.910171.110171.1

=

=

m

227.110227.1 10 == m For proton kgmm p

2710673.1 == and

27

2525

10673.110171.110171.1

=

=

m

02863.010863.2 12 == m For electron kgmm 2710695.6 == and

27

2525

10695.610171.110171.1

=

=

m

01431.010431.1 12 == m

Problem 2-27 Calculate the de Broglie wavelength of an alpha particle accelerated through 600 volts. Solution Now

JeVT 17190 10612.9)10602.1)(600( === The de Broglie wavelength in terms of kinetic energy is given by


Kmh

ph

2

==

m131727

34

1086.5)10612.9)(10645.6(2

10626.6

=

=

Problem 2-28 The wavelength of the yellow spectral emission line of sodium is 589 nm. At what kinetic energy would an electron have the same de Broglie wavelength? K.U. B.Sc. 2001, P.U. B.Sc. 2009 Solution The de Broglie wavelength is given by

Kmh

ph

02==

Square both sides

Kmh

0

22

2= or 2

0

2

2 mhK =

JK 252919234

10947.6)10589)(10109.9(2)10626.6(

=

=

eVeVK 61925

1034.410602.110947.6

=

=

Problem 2-29 If the de Broglie wavelength of a proton is 0.113 pm, (a) What is the speed of the proton and (b) through what electric potential would the proton have to be accelerated from rest to acquire this speed? Solution

(a) The de Broglie wavelength for proton is given by

vm

hp

= or pmh

v =


161227

34

10504.3)10113.0)(10673.1()10626.6(

=

= msv

(b) If Vo is the potential through which the proton has to be accelerated to acquire speed v, then

20 2

1vmeV p=

or )10602.1(2)10504.3)(10673.1(

2 1926272

0

==

e

vmV p

kVvoltsV 11.6410411.6 40 == Problem 2-30 A thermal neutron in a substance at temperature T is a neutron with kinetic energy equal to )2/3( kT , where

KJk /10381.1 23= is the Boltzmann constant. Determine the wavelength of a thermal neutron in a reactor at (a) 300 K and (b) 800 K. Solution The de Broglie wavelength is given by

mkTh

kTm

hmKh

ph

3232

2=

===

For a thermal neutron

TT

9

2327

34 10515.2)10381.1)(10675.1(3

)10626.6(

=

=

(a) T = 300 K , m109

10452.1300

10515.2

=

=

(b) T = 800 K , m119

10893.8800

10515.2

=

=


Problem 2-31 What is the de Broglie wavelength of an electron that is accelerated through a potential difference of 50 kV in a colour T.V. set? Solution The de Broglie wavelength is given by

mVeh

mKh

ph

22===

)10602.1)(1050)(10109.9(210626.6

19331

34

=

m1210485.5 =

Problem 2-32 An electron, a proton and a lead atom have the same de Broglie wavelength = 280 pm. Determine the kinetic energy of each. Solution The de Broglie wavelength in terms of kinetic energy K is given by

mKh

ph

2==

Squaring both sides

mKh

2

22

=

or mmm

hK48

212

234

2

2 10800.2)10280(2

)10626.6(2

=

==

For electron kgmm e3110109.9 == and

JK 183148

10074.310109.910800.2

=

= or 19.2 eV

For proton kgmm p2710673.1 == and

JK 212748

10674.110673.110800.2

=

= or 19.2 eV


For lead atom kgm 2510454.3 =

JK 242548

10107.810454.310800.2

=

= or 510061.5 eV

Problem 2-33 Calculate the kinetic energy of a neutron whose de Broglie wavelength is 10 -14 m. Solution The de Broglie wavelength in terms of kinetic energy K is given

by Km

hph

n2==

Squaring both sides Km

hn2

22

=

2

2

2 nmhK =

JK 1221427234

10311.1)101)(10675.1(2)10626.6(

=

=

MeVK 18.8= Q JeV 1910602.11 = As rest mass energy of a neutron is 939 MeV, therefore the use of non-relativistic expression is justified.

Problem 2-34 Calculate the de Broglie wavelength of an electron having kinetic energy equal to 2 MeV. Solution Now eVMeVK 61022 ==

JJK 13196 10204.3)10602.1(102( == The rest mass energy of the electron is Jcm 14283120 10187.8)10998.2)(10109.9( ==


The kinetic energy of electron is greater than its rest mass energy (about eight times), therefore the relativistic expression must be used i.e.

20cmKE +=

20

420

22 cmKcmcp +=+ since 420222

cmcpE += Square both sides

420

20

2420

22 2 cmcKmKcmcp ++=+

2

20

2

20

22 )2(2

c

cmKKc

cKmKp

+=

+=

c

cmKKp

)2( 20+=

Hence

)2( 20cmKKhc

ph

+==

)10187.8(2)10204.3){(10204.3()10998.2)(10626.6(

141313

834

+

=

m1310044.5 =

Problem 2-35 An electron has a de Broglie wavelength m10100.5 . (a) What is its momentum? (b) What is its speed? (c) What voltage was needed to accelerate it to this speed? B.U. B.Sc. 2009A Solution

(a) smkghp /10325.1100.510626.6 24

10

34

=

==

(b) vm

hph

0

==

smm

hv /10455.1)100.5)(10109.9(

10626.6 61031

34

0

=

==


(c) As 1005.010998.210455.1

8

6


2-2 TESTING DE BROGLIES HYPOTHESIS Problem 2-37 A neutron crystal spectrometer utilizes crystal planes of spacing d = 73.2 pm in a Beryllium. What must be the Bragg angle so that only neutrons of energy K = 4.2 eV are reflected? Consider only first-order reflections. Solution As neutrons carry no charge therefore the treatment of X-ray diffraction is equally applicable. According to Braggs law md =sin2

dm

2sin =

Now 1=m and mmpmd 1112 1032.7102.732.73 === ,

Kmh

n2=

m111927

34

10396.1)10602.12.4)(10675.1(2

)10626.6(

=

=

Hence 0954.0)1032.7)(2()10396.1)(1(

sin 1111

=

=

05.5=

Problem 2-38 A Davisson-Germer type experiment using nickel gives peak intensity in the reflected beam when the incident and reflected beams have an angular separation of 110o. What is the electron energy? (For nickel d = 0.091 nm) Solution According to Braggs law sin2dm = Now mmnmd 119 101.910091.0091.0 === ,

00

402

)100180(=

= and 1=m . Hence


mm

d 10011 1017.11

40sin)101.9(2sin2

=

==

The de Broglie wavelength in terms of kinetic energy K of the electron is given by

Kmh

e2=

or 21131

234

2

2

)1017.1)(10109.9(2)10626.6(

2

==

emhK

eVJK 1101076.1 17 ==

Problem 2-39 A beam of thermal neutrons from a nuclear reactor falls on a crystal of calcium fluoride, the beam direction making an angle with the surface of crystal. The atomic planes parallel to the crystal surface have an interplanar spacing of 54.65 pm. The de Broglie wavelength of electrons in the incident beam is 11.0 pm. For what value of will the first three orders of Bragg-reflected neutron beams occurs? Solution According to Braggs law md =sin2

dm

2sin =

Now 1=m , 10067.0)64.54(211

2sin ===

d , 636450 =

2=m , 20132.064.54

1122

sin ===d , 1563110 =

3=m , 30198.0)64.54(2)11(3

23

sin ===d , 6343170 =

Hence first three orders of Bragg-reflected neutron beams occur at angles 636450 , 1563110 and 6343170 .


2-3 HEISENBERG UNCERTAINTY PRINCIPLE Problem 2-40 What will be the mean time interval between production and decay of short lived particle of rest energy 3097 MeV, the uncertainty of measurement being only 0.063 MeV? P.U. B.Sc. 2001 Solution According to Heisenbergs Uncertainty Principle,

pi2))(( htE =

or )10602.110063.0(2)10626.6(

)(2 19634

=

=

pipi Eh

t

Q JeV 1910602.11 = st 2010045.1 = This value of time can be identified as the mean lifetime of the particle.

Problem 2-41 The energy of a certain nuclear state can be measured with an uncertainty of 1 eV. What is the minimum lifetime of this state? Solution According to Heisenbergs Uncertainty Principle,

pi2))(( htE =

or )10602.1(2)10626.6(

)(2 1934

=

=

pipi Eh

t

Q JeV 1910602.11 = st 1610583.6 =


Problem 2-42 A nucleus in an excited state will return to its ground state, emitting a gamma ray in the process. If its mean lifetime is 8.7 ps in a particular excited state of energy 1.32 MeV, find the uncertainty in the energy of the corresponding emitted gamma-ray photon. Solution The uncertainty in the energy of the emitted gamma-ray photon is given by the relation

Jt

hE 231234

10212.1)107.8(2)10626.6(

)(2

=

=

=

pipi

eVeVeVE 66.7510566.710602.110212.1 5

19

23

==

=

Problem 2-43 A pulsed ruby laser has an output of 2.0 GW (gigawatts) and produces a pulse of 10 psec duration. What is the uncertainty in the measurement of the laser energy? Solution According to Heisenbergs Uncertainty Principle,

pi2))(( htE =

or Jt

hE 231234

10055.1)1010(2)10626.6(

)(2

=

=

=

pipi

the desired uncertainty in the measurement of energy.

Problem 2-44 The life time of certain meson is about 2310 second. How accurately its rest energy be known? Solution According to Heisenbergs Uncertainty Principle,

pi2))(( htE =


)(2 thE

=pi

)101(2)10626.6(

23

34

=

piE

MeVorJE 86.6510055.1 11= the desired uncertainty in the measurement of energy.

Problem 2-45 An atom is in an excited state has a lifetime of 12 ns; in a second excited state the lifetime is 23 ns. What is the uncertainty in energy for a photon emitted when an electron makes a transition between these two states? Solution According to Heisenbergs Uncertainty Principle,

pi2))(( htE = or )(2 t

hE

=pi

nst 121 = , JhE )1012(2 91 = pi

nst 232 = , JhE )1023(2 92 = pi

Now

JhEEEtransition )102312(235

921 =+=

pi

J26934

10337.1)102312(2)10626.6(35

=

=

pi

eVeVEtransition9

19

26

10346.810602.110337.1

=

=

Problem 2-46 A microscope using photons is employed to locate an electron in an atom to within a distance of 12 pm. What is the minimum uncertainty in the momentum of the electron in this way?


Solution The minimum uncertainty minppx = in the momentum of electron is given by

2412

34

min 10788.8)1012(210626.6

)(2

=

=

=

pipi x

hp kg m s-1

Problem 2-47 A mass of 1 g has a speed of 1 cm/s. If its speed is uncertain by 1 percent, what is order of magnitude of minimum uncertainty in its position? Solution According to Heisenbergs Uncertainty Principle,

pi2))(( hxp =

vm

hp

hx

=

=

pipi 22 (1)

Now kgggm 96 1011011 === smscmv /101/1 2==

smvofv /101)101)(100/1(%1 42 === Substitute these values in Eq. (1) mx 2149

34

10)101)(101(210626.6

=

pi

Problem 2-48 A proton is confined to a space 1 fm wide (about the size of atomic nucleus). What is the minimum uncertainty in its velocity? Solution According to Heisenbergs Uncertainty Principle,

pi2))(( hxp =


pi2))}(({ hxvm p =

pi2))(( hxvm p =

smxm

hv

p

/10303.6)101)(10673.1(210626.6

)(26

1827

34

=

=

=

pipi

the desired uncertainty in velocity.

Problem 2-49 The position of an electron is measured to an accuracy of 1 . What is uncertainty in its velocity? Solution According to Heisenbergs Uncertainty Principle,

pi2))(( hxp =

pi2))}(({ hxvme =

pi2))(( hxvme =

)(2 xmh

ve

=pi

sm /10158.1)101)(10109.9(210626.6 6

1031

34

=

=

pi

the desired uncertainty in velocity of electron.

Problem 2-50 The position and momentum of a 1 keV electron are simultaneously determined. If its position is located to within 1 , what is the percentage of the uncertainty in its momentum?


Solution The momentum of the electron in terms of its kinetic energy is given by

)10602.1101)(10109.9(22 19319 == Kmp e Q

JeV 1910602.11 = 2310708.1 =p kg m s-1

The uncertainty in momentum is 24

10

34

10055.1)101(210626.6

)(2

=

=

=

pipi x

hp kg m s-1

Percentage uncertainty in momentum

100=pp

% = 10010708.110055.1

24

24

% = 6.18 %


CONCEPTUAL QUESTIONS (1) Why is the wave nature of matter not more apparent in our daily observations? Answer: - The wavelength associated with daily life objects is extremely small )10( 30 m and can not be measured experimentally. (2) If an electron and a proton travel at the same speed, which has the shorter wavelength? Answer: - The de Broglie wavelength is given by

vm

hph

== or p1

The proton will have shorter wavelength because it has more momentum than electron if both are moving with same speed. (3) Do you agree with the statement that electromagnetic radiation will show wave and particle aspect in the same experiment? Answer: - No, it will show one aspect only in an experiment. (4) If the wavelength of an electromagnetic wave is progressively reduced, what will happen to its particle behaviour? Answer: - The particle nature of electromagnetic radiation will become more and more dominant as its wavelength is progressively reduced. (5) What will be the effect on the wave nature of a particle if h = 0? Answer: - The particle will have no wave nature and can be described exclusively by classical mechanics. (6) If a proton and an electron has the same kinetic energy, which has the longer de Broglie wavelength? Explain. Answer: - The de Broglie wavelength in terms of kinetic energy K is given by

Kmh

ph

2==


For a given kinetic energy of the particle, its de Broglie wavelength becomes

m

1

As electron is lighter than proton, therefore it will have longer de Broglie wavelength. (7) Name an instrument based on de Broglie hypothesis. Answer: - Electron microscope. (8) How does the de Broglie wavelength associated with a particle compare with the dimensions of the particle? Answer: - The de Broglie wavelength is always smaller than the dimensions of the particle. (9) If the kinetic energy of a particle increases, then what will be the effect on its de Broglie wavelength? Answer: - The de Broglie wavelength in terms of kinetic

energy K is given by Km

hph

n2== . For a given particle we

get

K1

Hence a decrease in de Broglie wavelength of a particle will be observed with an increase in its kinetic energy. (10) If an electron has a wavelength, does it also have a colour? Explain. Answer: - An electron will exhibit wavelike properties if it possesses a kinetic energy of 150 eV. This energy corresponds to a wavelength of 1 which is in the ultraviolet region (more precisely X-ray region), therefore it has no visible colour. (11)Why Heisenbergs Uncertainty Principle is not significant at macroscopic level? Answer: - Due to small value of h i.e.

sJh == 3410055.12 pi

h


(12) Name an experiment which will confirm the wave-like behaviour of particles? Answer: - Davisson and Germer experiment (Diffraction of electrons by single crystals).


ADDITIONAL PROBLEMS

(1) Calculate the wavelength associated with a bullet having mass 10 grams and moving with a velocity of 500 m/s. B.U. B.Sc. 1991A

(2) A particle of mass one gram moves at a speed of 1 mm/s. Calculate the de Broglie wavelength associated with this particle. K.U. B.Sc. 1999

(3) Calculate the de Broglie wavelength associated with an electron after it has been accelerated from rest by a potential difference of 1 MV.

B.U. B.Sc.(Hons.) 1992A (4) Calculate the de Broglie wavelength of an electron

moving at a speed of 5101 m/s. (5) Calculate the de Broglie wavelength of 1 MeV alpha

particles. (6) Compare velocities and kinetic energies of an

electron and of a neutron for which the de Broglie wavelength is 1 .

Answers

(1) 3410325.1 m (2) m2810626.6 (3) 1210227.1 m (4) 7.273 nm (5) m41044.1 (6) ,/1026.7 6 sm sm /1096.3 3 , eVandeV 082.0015.0

chapter 2 wave nature of matter (pp 43-72)

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